\documentclass[a4paper]{article}
\begin{document}
\title{Homework 1}
\author{5120719013 Feng Shi}
\date{2013.06.03}
\maketitle

\section{Ex 2.5}
Solution:

When $d=2$, let the line through the center be $x-axis$,
let
	\begin{displaymath}
		S_{2}(x_{0}) = 2 \int_{0}^{x_{0}} \sqrt{1+{(\frac{-x}{\sqrt{2-x^{2}}})}^{2}}
		=2\sqrt{2} arcsin(x_{0})
	\end{displaymath}
	\begin{displaymath}
		S'_{2}(x_{0})=\frac{2 \sqrt{2}}{\sqrt{2-x_{0}^{2}}}
	\end{displaymath}
	$S'_{2}(x)$ is the projected surface area on the line.

	\vspace{6pt} 

When $d=3$, let the line through the center be $x-axis$,
let 
	\begin{displaymath}
		S_{3}(x_{0}) = \int_{0}^{x_{0}} 2\pi f(x) \sqrt{1+f'(x)^{2}} dx
	\end{displaymath}
	where $f(x)$ is the 3-dimensional sphere $f(x)=\sqrt{3-x^{2}}$.
	\begin{displaymath}
		S_{3}(x_{0}) = \int_{0}^{x_{0}} 2\pi \sqrt{3-x^{2}} \sqrt{1+\frac{x^{2}}{3-x^{2}}} dx 
		= \int_{0}^{x_{0}} 2\sqrt{3}\pi dx
	\end{displaymath}
	So the projected surface area is always $2\sqrt{3}\pi$ along the line.

	\vspace{6pt}

When d is large, use the following approach to get an approximation of the surface area 
of the sphere of radius $\sqrt{d}$ when $d$ is large:
randomly generate points with each coordinate according to Gaussian with unit variance.
As said in Chapter 2.5, those randomly generated points will lie in a thin annulus of width $O(1)$
at radius $\sqrt{d}$. So this annulus can be an approximation of the surface area 
of the sphere of radius $\sqrt{d}$ when $d$ is large.

As each coordinate of the points we generate is Gaussian, 
we will get a unit variance Gaussian if we
project the annulus onto any one axis through the origin.
So we will get a unit variance Gaussian if we project the surface area 
of the sphere of radius $\sqrt{d}$
onto a line through the center when $d$ is large.
So for large $d$ the projected surface area should
 behave like a Gaussian.

 \vspace{6pt}

\section{Ex 2.10}
Solution:

\begin{displaymath}
	V(d) = \frac{d}{2} \frac{\pi ^ {\frac{d}{2}} }{\Gamma( \frac{d}{2})}
	= \frac{\pi ^ {\frac{d}{2}}}{\Gamma(1 + \frac{d}{2})}
	= \left\{ \begin{array}{ll}
			\frac{\pi^{\frac{d}{2}}}{(d/2)!} & \textrm{if $d$ is even}\\
			\frac{ \pi ^ {\lfloor d/2 \rfloor} 2 ^ {\lceil d/2 \rceil}}{d!!} & \textrm{if $d$ is odd}
	\end{array} \right.
\end{displaymath}

so we have the recursion:

$V(0) = 1$

$V(1) = 2\pi$

$V(d) = \frac{2\pi}{d} V(d-2)$ for $d>1$

so we can calculate:

$V(0) = (1/0!)\pi^{0}=1$

$V(1) = (2^{1}/1!!)\pi^{0}=2$

$V(2) = (1/1!)\pi^{1}=\pi$

$V(3) = (2^{2}/3!!)\pi^{1}=\frac{4}{3}\pi$

$V(4) = (1/2!)\pi^{2}=\frac{1}{2}\pi^{2}$

$V(5) = (2^{3}/5!!)\pi^{2}=\frac{8}{15}\pi^{2}$

$V(6) = (1/3!)\pi^{3}=\frac{1}{6}\pi^{3}$

for $d>6$, $\frac{2\pi}{d}<1$, so the volumes will be decreasing.
Thus we can see when $d=5$, the volume of d-dimensional unit sphere is maximum.
\end{document}
